Ultraviolet A 1426-discrete square roots (Extended Euclidean)

Link to the question: Ultraviolet A 1426-discrete square roots

Returns X, N, R, and returns all satisfied R, so that R2 returns X % N, and R is a solution.

Solution:

• R2? R2 = K? N
• (R? R) (R + R) = K? N
• => AA = (R + r), BB = (r? R), A, and B are the factors of N.
  So if we enumerate a and B, r = r? AA = BB? R
• AA + BB = 2? R
  Expand the Euclidean solution and free all satisfied items into the set to automatically remove duplicates.

#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <algorithm>using namespace std;typedef long long ll;ll X, N, R;set<ll> ans;void exgcd (ll a, ll b, ll& d, ll& x, ll& y) {    if (b == 0) {        d = a;        x = 1;        y = 0;    } else {        exgcd (b, a%b, d, y, x);        y -= (a/b) * x;    }}void solve (ll A, ll B, ll C) {    ll d, x, y;    exgcd(A, B, d, x, y);    if (C % d)        return;    x = x * C / d;    x = (x % (B/d) + (B/d))%(B/d);    ll s = x * A - C / 2;    ll k = A * B / d;    for (ll i = s; i < N; i += k)        if (i >= 0)            ans.insert(i);}int main () {    int cas = 1;    while (scanf("%lld%lld%lld", &X, &N, &R) == 3 && X + N + R) {        ans.clear();        ll m = sqrt(N+0.5);        for (ll i = 1; i <= m; i++) {            if (N%i)                continue;            solve(i, N/i, 2*R);            solve(N/i, i, 2*R);        }        printf("Case %d:", cas++);        for (set<ll>::iterator i = ans.begin(); i != ans.end(); i++)            printf(" %lld", *i);        printf("\n");    }    return 0;}